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Phantasy [73]
2 years ago
13

For the common-emitter, common-base and emitter-follower amplifier designs,what is the primary benefit of each amplifier?

Physics
1 answer:
kati45 [8]2 years ago
6 0

Answer:

Explanation:

A common emitter amplifier works by inverting. It does have low input impedance. Despite its low input impedance, it poses an otherwise high output impedance.

The common base circuit performs optimally when it acts as a current buffer. It has the ability to take an input current at a low input impedance, and transmit almost the same current to an impedance with a higher output

The primary benefit of emitter follower amplifier is that the transistor is able to provide current and power gain. Although this transistor takes in little current from the input. It still provides an impedance with a low output to a circuit by exercising the output of the follower. This then translates to that the output under load not dropping.

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Help please correct answer i will mark brainliest​
murzikaleks [220]

Answer:

19.21ms-¹

Explanation:

that is the solution above

6 0
3 years ago
How much is the velocity of a body when it travels 600m in 5 minutes?( with the full process)​
Tcecarenko [31]
Its average speed, pretending that it traveled at a constant speed, is

v = s / t

= 600 m

5 x 60 s

= 2 m/s

but to be a velocity it needs a direction as well as a speed.

( Sorry. Can’t find a division line to put between the 600 m and the 5 x 60 s )
6 0
3 years ago
Orbitals that are of equal energy may be referred to as.
ivanzaharov [21]

Answer:

degenerate

Explanation:

5 0
2 years ago
When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the
dlinn [17]

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

F_1 = F \frac{{343}}{(343-V)}

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

F_2 = F\frac{343}{(343+V)}

We are also so told the perceived frequency decreases by 11.9%

F_2 = F_1 - 9.27\% \text{ of } F_1

F_2 = F_1-0.0927F_1

F_2 = 0.9073F_1

Equating,

F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})

\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}

0.9073(343+V) = 343-V

(0.9073)(343)+(0.9073)V = 343-V

V+0.9073V = 343-(0.9073)(343)

Solving for V,

V = 16.67 m/s

5 0
3 years ago
Examples of stored energy
jeyben [28]

Answer:

Biomass, petroleum, natural gas, and propane are examples of stored chemical energy. Energy is energy stored in objects by the application of a force

3 0
3 years ago
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